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Statement:

Assume $V$ is a vector space over a field $K$. Let $\phi: V \rightarrow V$ be a linear map.

Then $\lambda \in K$ is an eigenvalue of $\phi$ iff $\lambda$ is a root of the characteristic polynomial $P$.

Proof:

Let $M$ be the matrix of $\phi$ and let $\lambda \in K$ be arbitrary.

We have $P(\lambda) = \det(\lambda E_n - M) = 0$ iff $\lambda Id_V - \phi$ is not bijective (and not injective). This follows from the theorems ...

This is equivalent to

$Eig_\lambda(\phi) = kern(\lambda Id_V - \phi) \neq 0$, which means that the eigenspace of $\lambda$ is not the nullspace. Therefore, $\lambda$ must be an eigenvalue of $\phi$.

The theorems he mentioned are the following:

Let $M$ be a $n \times n$ matrix. Then the following statements are equivalent:

  1. $det M \neq 0$.
  2. The rows of $M$ are linear indepentent.
  3. $M$ is invertible.
  4. $rank(M) = n$.

and

Assume $V$ and $W$ are vector spaces over $K$ with $dim(V) = n$ and $dim(W) = m$. Let $\phi: V \rightarrow W$ be a linear map that can be denoted by the matrix $M$. Then the following statements are true:

  1. $\phi$ is injective iff the columns of the matrix are linear independent.
  2. $\phi$ is surjective iff the columns of the matrix generate $K^m$.
  3. If $m = n$, then $\phi$ is bijective iff the columns of the matrix are a base of $K^m$, and this holds iff $M$ is invertible.

Questions

My question is: How do we know that $\phi$ is not injective given these two theorems and the argumentation above?

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Julian
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  • Linear operators on finite-dimensional vector spaces are injective iff they are surjective. – Math1000 Oct 08 '16 at 10:11
  • So by concluding that $\phi$ is not bijective, we also conclude that $\phi$ mustn't be either injective or surjective or both, but already concluding one of those properties already gives us the other one? – Julian Oct 08 '16 at 10:15
  • If $\dim V=n$ and $T:V\to V$ is a linear map, then by the rank-nullity theorem, $$\dim T(V) + \dim \ker(T) = n. $$ $T$ is injective iff $\dim\ker(T)=0$, and $T$ is surjective iff $\dim T(V)=n$. – Math1000 Oct 08 '16 at 10:35

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